1 Lec 26: Frictionless flow with work, pipe flow

1

Lec 26: Frictionless flow with work, pipe flow

2

• For next time:– Read: § 8-14 to 8-18– HW 13 due Monday, December 1, 2003

• Outline:– Bernoulli equation in pipe flow– Accounting for pumps and turbines– Accounting for friction in pipes

• Important points:– Don’t forget the conservation of mass equation– Be careful how you account for pump and

turbine losses– Understand the Moody Diagram

3

Fluid Mechanics

• A venturi tube or meter--converging-diverging nozzle frequently used to measure the volumetric flowrate of a fluid. It must be inserted into a pipe or duct as a part of the pipe or duct.

1 2

4

TEAMPLAYTEAMPLAY

• For a venturi such as that shown before, the following data apply: dia1 = 6.0 in,

dia2 = 4.0 in. The pressure difference

P1 – P2 = 3 psi. Water with a density of

62.4 lbm/ft3 is flowing. Find the rate of flow in ft3/min.

• Hint: use flowrate Qv = A1V1 = A2V2

5

Pipe flow

• We previously had the Bernoulli equation in this form and we observed that the terms are energy per unit mass.

• In a pipe (or duct), this equation represents the mechanical energy per unit mass at a flow cross section.

constant

2

2P

gz

v

6

Pipe flow

• For frictionless flow, the mechanical energy will be the same at every cross section of the pipe and

2

22

1

21 gz

2

Pgz

2

P

21 vv

7

Pipe flow

Cross section 1

Cross section 2

8

Pipe flow

• In real pipe or duct flows, energy must be used to overcome friction, and so at subsequent cross sections the energy is less.

• Pipe flow--examples are water systems and petroleum pipeline systems.

• Duct flow--examples are air conditioning ducts in buildings.

9

Pipe and duct flow

• We had the following equation for incompressible steady flow:

• or, replacing v by and rearranging

12

21

22

1212 22)()( zzguuPPvwq

vv

)(22 1212

21

2212 uuzzg

PPwq

vv

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Pipe and duct flow

)()( 1212 TTcuu v •The term represents the increase in internal energy that occurs due to friction as the fluid flows in a pipe or duct. The textbook calls this a mechanical loss term, emech,

loss.

•The text also splits the work term into two:

•wpump represents pumping power input or the power required by a pump or fan to move the flow.

•Wturbine represents the work done by a water turbine, for example.

11

Pipe and duct flow

• Thus the book arrives at the following equation for adiabatic flow:

• or

)(22 1212

21

2212 uuzzg

PPw

vv

lossmechturbine

shaftpump

ewgzP

wgzP

,2

222

,1

211

2

2

v

v

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Pipe and duct flow

• The previous equation

• is equation 11-31 in the text book, where the equation (incorrectly) appears thusly

lossmechturbineupump ewgz

Pwgz

P,2

222

,1

211

22

vv

lossmechturbineshaftpump ewgz

Pwgz

P,2

222

,1

211

22

vv

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Pipe and duct flow

• The shaft work includes the work to increase the mechanical energy of the fluid and to overcome losses. Eq 11-31 and 11-32 should read:

• where if we forget the wturbine for the moment, wpump,u raises the mechanical energy of the flow.

turbineupump wgz

Pwgz

P 2

222

,1

211

22

vv

turbineupump WgzP

mWgzP

m

2

222

,1

211

22

vv

14

Pipe and duct flow

• Divide the previous equation by g to get an equation in terms of head:

lossmech

turbine

shaftpump

g

e

gz

gg

P

gz

gg

P

,

2

222

,1

211

w

2

w

2

v

v

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Pipe and duct flow

• The term (e/g)mech,loss is identified as the head loss term, and we calculate it later.

L,

Hlossmechg

e

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Pipe and duct flow

• Not all of the shaft power going into the pump is converted into useful mechanical energy supplied to the fluid, such as raising water up into a water tower.

• Some is lost in frictional heating that manifests itself as a slight temperature rise of the fluid.

• Thus wpump,shaft= wpump,u+emech,losses

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Pipe and duct flow

• This loss gives rise to the following pump efficiency

in shaft,

upump,ump w

w

input work Mechanical

fluid the tosuppliedenergyMechanical

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TEAMPLAYTEAMPLAY

• Consider a water pump in a horizontal, constant area pipe. The mass flow rate is 50 kg/s, and the pump receives 17.0 kW of power from its driver. The delta P (pressure change) across the pump is 250 kPa.

• Determine the mechanical efficiency of the pump and the temperature rise of the fluid.

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Pipe and duct flow

L2

22

1

21 Hz

g2g

Pz

g2g

P

21 vv

•Using the equation in terms of head loss and omitting the work terms (meaning we do not have a pump, turbine or compressor in the system),

•Where the previous equation in terms of flow energy per unit mass has been divided by g (not gc) to yield units of head (length).

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Pipe and duct flow

• It can be shown that the the head loss term can be determined in terms of the shear stress at the wall, o.

• Where is the length of the pipe and D is the diameter of the pipe.

oL gD

4H

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Pipe and duct flow

• The shear stress at the wall is a function of five independent variables.

o = o(,,V,D,e) is the dynamic or absolute viscosity• V is the mean velocity of the flow• e (or ) is the surface roughness of the

pipe, and can be described as a length.

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Pipe and duct flow

• With five independent variables, the solution is very complex.

• However, the complexity is reduced to two variables by use of only two non-dimensional variables.

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Pipe and duct flow

• The non-dimensional shear stress can be expressed as

D

e,f

2

2o VD

V

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Pipe and duct flow

VD

= Re, the Reynolds number (non- dimensional)

e/D = relative roughness

= ratio of inertia forces/viscous forces

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Pipe and duct flow

•The non-dimensional shear stress is tabulated as a function of the Reynolds number and the surface roughness.

•The non-dimensional shear stress is called a friction factor.

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Pipe and duct flow

• Fanning friction factor, , used in heat transfer:

• Moody friction factor, f, used in fluid mechanics:

2

2o

V

4f

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Pipe and duct flow

• Moody friction factor, f, used in fluid mechanics is typically given as

• and appears on page 974 of the textbook.

2o8

4fV

2

4 2o

V

28

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Pipe and duct flow

• So,

g2D

f8

f

gD

4

gD

4H

22

oL

VV

•And

D

eRe,offunctionf

30

T hus

L2

22

1

21 Hz

g2g

Pz

g2g

P

21 vv

gDf

2

2V

P i p e a n d d u c t fl o w

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TEAMPLAYTEAMPLAY

• Determine the pressure drop over 1000 ft of pipe carrying water if the flowrate is 70 cfs and the pipe diameter is 30 inches. Assume that the pipe has a roughness, e, of 0.1 inches and the temperature is 25 C. Water has a density of 62.4 lbm/cubic foot.

32

TEAMPLAYTEAMPLAY

A 2 ft diameter cylindrical aire conditioning duct hands horizontally from a ceiling and carries 10,000 CFM of 55 F air. Relative roughness factor e/D = 0.0002. Find the pressure drop der 100 ft of duct length in the units of inches of water if 27.7 in of water is 1 psia.

1. Write and simplify the basic equation.2. Find the density, velocity of flow, Reynold’s

number and friction factor.3. Solve the problem.